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Thread: ddifferentiation using l'Hopitals rule

  1. #1
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    differentiation using l'Hopitals rule

    Hello, I am trying to diffferentiate using l'Hopital's rule:
    (e^(5+h)-e^5) / (h)
    to find the limit of h tending towards zero.

    I differentiate the numerator and denominator to get:

    (e^(5+h) - e^5) / (1)

    but I think I should be getting (e^(5+h)) / (1)
    Can someone please tell me how the second e^5 is done away with ?

    Thanks kindly.
    Last edited by fran1942; Apr 3rd 2012 at 05:39 PM.
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  2. #2
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    Re: ddifferentiation using l'Hopitals rule

    You don't differentiate using L'Hospital's Rule, you differentiate the top and the bottom because that's what you need to do to USE L'Hospital's Rule. Anyway

    $\displaystyle \displaystyle \begin{align*} \lim_{h \to 0}\frac{e^{5 + h} - e^5}{h} &= \lim_{h \to 0}\frac{e^5e^h - e^5}{h} \\ &= \lim_{h \to 0}\frac{e^5\left(e^h - 1\right)}{h} \\ &= e^5\lim_{h \to 0}\frac{e^h - 1}{h} \textrm{ which goes to }\frac{0}{0}\textrm{ so L'Hospital's Rule can be used} \\ &= e^5\lim_{h \to 0}\frac{\frac{d}{dx}\left(e^h - 1\right)}{\frac{d}{dx}\left(h\right)} \textrm{ by L'Hospital's Rule} \\ &= e^5\lim_{h \to 0}\frac{e^h}{1} \\ &= e^5\lim_{h \to 0}e^h \\ &= e^5 \cdot 1 \\ &= e^5 \end{align*}$
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  3. #3
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    Re: differentiation using l'Hopitals rule

    Quote Originally Posted by fran1942 View Post
    Hello, I am trying to diffferentiate using l'Hopital's rule:
    (e^(5+h)-e^5) / (h)
    to find the limit of h tending towards zero.

    I differentiate the numerator and denominator to get:

    (e^(5+h) - e^5) / (1)

    but I think I should be getting (e^(5+h)) / (1)
    Can someone please tell me how the second e^5 is done away with ?

    Thanks kindly.

    Circular reasoning
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    Re: ddifferentiation using l'Hopitals rule

    e^5 is a constant and differentiates to zero
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